Write $z_i(x)=\langle a_i,x\rangle+b_i$. The <log-sum-exp function> is convex and composition with the <affine functions> $z_i$ preserves convexity, so $f_\beta$ is <convex>. Directly, its <Hessian matrix> will also be shown <positive semidefinite matrix>[positive semidefinite] in part d.

Let $M=f(x)=\max_i z_i(x)$. Factoring $e^{\beta M}$ out of the sum gives
$$
f_\beta(x)
=M+\frac1\beta\log\sum_i e^{\beta(z_i(x)-M)}.
$$
At least one term in the sum is $1$, while every term is at most $1$. Therefore
$$
1\leq\sum_i e^{\beta(z_i-M)}\leq m,
$$
and hence
$$
\boxed{f(x)\leq f_\beta(x)\leq f(x)+\frac{\log m}{\beta}.}
$$
Thus $f_\beta$ is a uniform <smooth maximum> of the affine pieces of $f$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 339 / 1 / c / Solution

Composing the <log-sum-exp function> with finitely many <affine functions> gives a differentiable approximation to their pointwise maximum. Increasing the inverse-temperature parameter $\beta$ reduces the uniform approximation error while increasing the <Lipschitz gradient> constant.

Smooth maximum

{title2=$\operatorname{LSE}_\beta$}
{wiki}

For $z\in\mathbb R^m$ and $\beta>0$, the scaled log-sum-exp function is
$$
\operatorname{LSE}_\beta(z)
=\frac1\beta\log\sum_{i=1}^m e^{\beta z_i}.
$$
It is a smooth <convex function> satisfying
$$
\max_i z_i\leq\operatorname{LSE}_\beta(z)
\leq\max_i z_i+\frac{\log m}{\beta}.
$$

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 Incoming links: Log-sum-exp function